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2 years ago

Please help! thank you~

Mathematics

1 answer:

DanielleElmas [232]2 years ago

3 0

Answer:

i would think u should add them together and that would be 21 but maybe you subtract the two years ago and get 19

Step-by-step explanation:

You might be interested in

$1000 is invested at 2.9% annual interest for 6 months.<br> a. Simple Interest b. Compound Interest

kirza4 [7]

Answer:

a. Simple interest = $174

b. Compund interest = $1187.1

Step-by-step explanation:

Given the following data;

Principal, P = $1000

Interest, R = 2.9%

Time, T = 6 months

a. To find the simple interest;

S.I = (PRT)/100

S.I = (1000*2.9*6)/100

S.I = 17400/100

S.I = $174

b. To find the compound interest;

$A = P(1 + \frac{r}{100})^{t}$

Substituting into the formula, we have;

$A = 1000(1 + \frac{2.9}{100})^{6}$

$A = 1000(1 + 0.029)^{6}$

$A = 1000(1.029)^{6}$

$A = 1000(1.1871)$

A = $1187.1

8 0

2 years ago

What is the least common multiple of15&25

nikklg [1K]

The LCM of 15 and 25 is 75.

3 0

2 years ago

If P(E) = 0.32, then what are the odds against E?

steposvetlana [31]

Answer:

0.68 : 0.32

Step-by-step explanation:

the probability of all the outcomes of an event is always 1 .

if P(E) = 0.32,

P(E') = 1 - 0.32

= 0.68

N.B that P(E') is the same as P( not E ).

since your answer should be given in the ratio form, as the odds against E ( probability not E : probability E),

P(E') : P(E)

0.68 : 0.32

hope this helps you!

-s.

7 0

1 year ago

67,802. the 0 has what place value?

natima [27]

Answer:

the 0 is in the tens place.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

2 years ago